Difference between revisions of "1992 AIME Problems/Problem 4"
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== Problem == | == Problem == | ||
− | In Pascal's Triangle, each entry is the sum of the two entries above it. | + | In Pascal's Triangle, each entry is the sum of the two entries above it. The first few rows of the triangle are shown below. |
− | == Solution == | + | <cmath>\begin{array}{c@{\hspace{8em}} |
+ | c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{4pt}}c@{\hspace{2pt}} | ||
+ | c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{3pt}}c@{\hspace{6pt}} | ||
+ | c@{\hspace{6pt}}c@{\hspace{6pt}}c} \vspace{4pt} | ||
+ | \text{Row 0: } & & & & & & & 1 & & & & & & \\vspace{4pt} | ||
+ | \text{Row 1: } & & & & & & 1 & & 1 & & & & & \\vspace{4pt} | ||
+ | \text{Row 2: } & & & & & 1 & & 2 & & 1 & & & & \\vspace{4pt} | ||
+ | \text{Row 3: } & & & & 1 & & 3 & & 3 & & 1 & & & \\vspace{4pt} | ||
+ | \text{Row 4: } & & & 1 & & 4 & & 6 & & 4 & & 1 & & \\vspace{4pt} | ||
+ | \text{Row 5: } & & 1 & & 5 & &10& &10 & & 5 & & 1 & \\vspace{4pt} | ||
+ | \text{Row 6: } & 1 & & 6 & &15& &20& &15 & & 6 & & 1 | ||
+ | \end{array}</cmath> | ||
+ | In which row of [[Pascal's Triangle]] do three consecutive entries occur that are in the ratio <math>3 :4 :5</math>? | ||
+ | |||
+ | == Solution 1== | ||
Consider what the ratio means. Since we know that they are consecutive terms, we can say | Consider what the ratio means. Since we know that they are consecutive terms, we can say | ||
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<cmath> 28n - 35 = 27n+27 </cmath> | <cmath> 28n - 35 = 27n+27 </cmath> | ||
<cmath> n = 62 </cmath> | <cmath> n = 62 </cmath> | ||
− | We can also evaluate for <math>k</math>, and find that <math>k = \frac{3(62+1)}{7} = 27.</math> Since we want <math>n</math>, however, our final answer is <math>\boxed{062.}</math> ~LaTeX by ciceronii | + | We can also evaluate for <math>k</math>, and find that <math>k = \frac{3(62+1)}{7} = 27.</math> Since we want <math>n</math>, however, our final answer is <math>\boxed{062.}</math> ~<math>\LaTeX</math> by ciceronii |
+ | |||
+ | ==Solution 2== | ||
+ | Call the row <math>x=t+k</math>, and the position of the terms <math>t-1, t, t+1</math>. Call the middle term in the ratio <math>N = \dbinom{t+k}{t} = \frac{(t+k)!}{k!t!}</math>. The first term is <math>N \frac{t}{k+1}</math>, and the final term is <math>N \frac{k}{t+1}</math>. Because we have the ratio <math>3:4:5</math>, | ||
+ | |||
+ | <math>\frac{t}{k+1} = \frac{3}{4}</math> and <math>\frac{k}{t+1} = \frac{5}{4}</math>. | ||
+ | |||
+ | <math>4t = 3k+3</math> and <math>4k= 5t+5</math> | ||
+ | |||
+ | <math>4t-3k=3</math> | ||
+ | <math>5t-4k=-5</math> | ||
+ | |||
+ | Solve the equations to get <math> t= 27, k=35</math> and <math>x = t+k = \boxed{062}</math>. | ||
+ | |||
+ | -Solution and LaTeX by jackshi2006, variables and algebra simplified by oinava | ||
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[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
− |
Latest revision as of 13:13, 20 April 2024
Problem
In Pascal's Triangle, each entry is the sum of the two entries above it. The first few rows of the triangle are shown below.
In which row of Pascal's Triangle do three consecutive entries occur that are in the ratio ?
Solution 1
Consider what the ratio means. Since we know that they are consecutive terms, we can say
Taking the first part, and using our expression for choose , Then, we can use the second part of the equation. Since we know we can plug this in, giving us We can also evaluate for , and find that Since we want , however, our final answer is ~ by ciceronii
Solution 2
Call the row , and the position of the terms . Call the middle term in the ratio . The first term is , and the final term is . Because we have the ratio ,
and .
and
Solve the equations to get and .
-Solution and LaTeX by jackshi2006, variables and algebra simplified by oinava
1992 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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All AIME Problems and Solutions |
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